The previous page, measures and measure spaces, set out the rules of the game: a measure is a countably additive set function on a σ-algebra, sending \varnothing to 0. But it took the existence of such a thing on \mathbb{R} — one that assigns [a,b] its length b-a — entirely on faith. This page pays that debt. We are going to build Lebesgue measure on the line from scratch, and the whole construction rests on a single primitive fact you have known since childhood: the length of an open interval (a,b) is b-a. Nothing else is assumed.

The plan has two movements. First we define a crude size — the outer measure \mu^{*} — that is defined for every subset of \mathbb{R} by wrapping it in intervals as tightly as possible. It is generous and well-behaved, but it fails the one law a measure must obey: it is only subadditive, not additive. Then Lebesgue's successor Carathéodory supplies the master stroke — a razor that carves out exactly the sets on which \mu^{*} is additive. On that family, and only there, \mu^{*} becomes an honest, complete measure. That is Lebesgue measure.

The outer measure: wrap it in intervals

Take any set A \subseteq \mathbb{R} — literally any set, no assumptions. A countable cover of A by open intervals is a sequence (a_1,b_1), (a_2,b_2), \dots whose union contains A. Every such cover has a total length \sum_n (b_n - a_n), a number in [0,\infty]. The Lebesgue outer measure of A is the cheapest cover you can find — the infimum of that total length over all admissible covers:

Two features are the whole point. First, this \inf always exists: the set of achievable total lengths is a nonempty set of non-negative reals (nonempty because \mathbb{R} = \bigcup_n (-n, n) is one cover, though a useless one for a bounded set), and every nonempty set of reals bounded below has an infimum by the completeness of \mathbb{R}. There is no risk that "the best cover" fails to exist — completeness guarantees a greatest lower bound even when no single cover attains it. Second, \mu^{*} is defined for every subset of \mathbb{R}, no exceptions — the very thing a genuine measure could not manage on all of \mathbb{R}. That universality is bought at a price we will discover shortly.

What outer measure gets right

Straight from the definition, three properties fall out, and one crucial property does not.

Empty set and monotonicity

\mu^{*}(\varnothing) = 0: the single interval (-\varepsilon, \varepsilon) covers \varnothing with length 2\varepsilon, and \varepsilon is arbitrary, so the infimum is 0. And \mu^{*} is monotone: if A \subseteq B then every cover of B is a cover of A, so the infimum for A is taken over a larger set of covers and can only be smaller — \mu^{*}(A) \le \mu^{*}(B).

Countable subadditivity — the \varepsilon/2^{n} trick

The workhorse. For any sequence of sets A_1, A_2, \dots (disjoint or not),

Proof. Fix \varepsilon > 0. By definition of the infimum, each A_n has a cover by open intervals whose total length overshoots \mu^{*}(A_n) by less than an amount we get to choose — and we choose the shrinking budget \varepsilon / 2^{n}:

Now pool all these intervals into one countable family (a countable union of countable sets is countable). It covers \bigcup_n A_n, so its total length bounds the outer measure of the union:

The geometric budget \sum_n \varepsilon/2^{n} = \varepsilon is the entire trick: infinitely many small errors sum to one small error. Since \varepsilon was arbitrary, the inequality holds without it. This one lemma underwrites everything below.

What outer measure gets WRONG

Here is the catch. \mu^{*} is not countably additive on all subsets — for disjoint sets the inequality above can be strict. That is not a defect we will patch; it is a theorem. The Vitali construction exhibits disjoint sets whose outer measures sum to more than the outer measure of their union. So \mu^{*} is an outer measure — monotone, subadditive, zero on \varnothing — but not yet a measure. Turning subadditivity into additivity is the job of Carathéodory's razor.

It really is length: \mu^{*}([a,b]) = b-a

A construction that failed to give an interval its own length would be worthless. It doesn't fail — but the proof has a surprising asymmetry.

The easy direction, \mu^{*}([a,b]) \le b-a. The single open interval (a-\varepsilon,\, b+\varepsilon) covers [a,b] and has length b-a+2\varepsilon. So the infimum is at most b-a+2\varepsilon for every \varepsilon>0, hence at most b-a. One cover was enough.

The hard direction, \mu^{*}([a,b]) \ge b-a, is where the real content lives, and it needs compactness. Take any open cover [a,b] \subseteq \bigcup_n (a_n,b_n). Because [a,b] is closed and bounded, the Heine–Borel theorem hands us a finite subcover. A short induction on a finite list of intervals covering an interval shows their lengths must sum to at least b-a (walk from a to b, at each step the interval containing the current point reaches strictly past it). So every cover — via its finite subcover — has total length \ge b-a, and therefore so does the infimum. Compactness is what stops an infinite cover from cheating with cancellation; without it the inequality is false.

And countable sets vanish: \mu^{*}(\mathbb{Q}) = 0

Enumerate a countable set as x_1, x_2, x_3, \dots (the rationals, say). Given \varepsilon>0, wrap x_n in the open interval \bigl(x_n - \tfrac{\varepsilon}{2^{n+1}},\, x_n + \tfrac{\varepsilon}{2^{n+1}}\bigr), of length \varepsilon/2^{n}. These intervals cover the whole set and their total length is

The cover has total length \varepsilon, arbitrarily small, so \mu^{*}(\mathbb{Q}) = 0. Every countable set is \mu^{*}-null. This is the rigorous version of the hand-wave on the Riemann page: \mathbf{1}_{\mathbb{Q}} "ought to" integrate to 0 because \mathbb{Q} is negligible — and now "negligible" has an exact meaning, outer measure 0. Notice the \varepsilon/2^{n} budget again: the same geometric series that gave subadditivity dissolves any countable set.

See the cover shrink

Drag the ε slider and watch the total length of the cover fall toward 0; drag k to add more points and see that even covering more of the set costs almost nothing, because interval number n has width \varepsilon/2^{n}. The readout is the exact finite sum \sum_{n=1}^{k} \varepsilon/2^{n} = \varepsilon\,(1 - 2^{-k}), always strictly below \varepsilon. Push ε down and the whole scaffold of intervals collapses to the points — that limit, total length 0, is the outer measure of a countable set.

For a solid interval [a,b] the picture would refuse to collapse: no matter how cleverly you cover it, compactness pins the total length at \ge b-a. Countable dust shrinks to nothing; a continuum keeps its length. That contrast is the entire difference between measure zero and positive measure.

Carathéodory's razor: which sets are measurable

We need a test that selects the sets on which \mu^{*} behaves additively — and the test must use nothing but \mu^{*} itself, since that is all we have. Carathéodory's idea is startlingly clean: a set is measurable if it splits every other set additively.

Read A as a knife. For any "test set" E, the knife cuts it into the part inside A and the part outside; A is measurable exactly when those two pieces' outer measures always add back to the whole. By subadditivity the \le direction is automatic (E = (E\cap A) \cup (E\cap A^{c})), so the condition really only asks for the reverse inequality \mu^{*}(E) \ge \mu^{*}(E\cap A) + \mu^{*}(E\cap A^{c}): the knife must not lose any measure. Sets that fail this — like the Vitali set — leak: their two pieces overstate the whole.

We state this without the full proof (the σ-algebra closure argument is a careful but elementary exercise in applying the splitting condition to nested test sets). Two corollaries deserve a spotlight. Because every open interval is measurable and \mathcal{M} is a σ-algebra, \mathcal{M} swallows the whole Borel hierarchy — \mathcal{B} \subseteq \mathcal{M}. And because \mu is complete, every set of outer measure zero is measurable: if \mu^{*}(A)=0 then for any test set E, monotonicity forces \mu^{*}(E\cap A)=0 and \mu^{*}(E\cap A^{c}) \le \mu^{*}(E), so the splitting condition holds trivially. Null sets pass Carathéodory's test for free.